Simplify the following expression: $n = \dfrac{9q^2 + 126q + 405}{q + 5} $
Explanation: First factor the polynomial in the numerator. We notice that all the terms in the numerator have a common factor of $9$ , so we can rewrite the expression: $ n =\dfrac{9(q^2 + 14q + 45)}{q + 5} $ Then we factor the remaining polynomial: $q^2 + {14}q + {45} $ ${5} + {9} = {14}$ ${5} \times {9} = {45}$ $ (q + {5}) (q + {9}) $ This gives us a factored expression: $\dfrac{9(q + {5}) (q + {9})}{q + 5}$ We can divide the numerator and denominator by $(q - 5)$ on condition that $q \neq -5$ Therefore $n = 9(q + 9); q \neq -5$